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in Arithmetic Progression by (43.0k points)
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Which of the following series is an arithmetic progression ? 

A) 2, 4, 6, 8, 

B) 1, 2, 4, 8, 

C) 4, 9, 16, 25, 

D) 3, 9, 12, 18,

2 Answers

+1 vote
by (57.0k points)
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Best answer

Correct option is A) 2, 4, 6, 8, 

(A) 2, 4, 6, 8, ......

\(a_2-a_1\) \(=4-2=2\)

\(a_3-a_2\) \(=6-4=2\)

and \(a_4-a_3\) \(=8-6=2\)

\(\because\) \(a_4-a_3\) \(=a_3-a_2\) \(=a_2-a_1\) \(=2\)

\(\therefore\) 2, 4, 6, 8, ...... is an arithmetic progression.

(B) 1, 2, 4, 8, ........

\(a_2-a_1\) \(=2-1=1,\)

\(a_3-a_2\) \(=4-2=2\)

\(\because\) \(a_3-a_2\) \(\neq a_2-a_1\)

\(\therefore\) 1, 2, 4, 8, ........ will not form an A.P.

(C) 4, 9, 16, 25, .......

\(a_2-a_1\) \(=9-4=5,\)

\(a_3-a_2\) \(=16-9=7\)

\(\because\) \(a_3-a_2\) \(\neq a_2-a_1\)

\(\therefore\) 4, 9, 16, 25, ....... will not form an A.P.

(D) 3, 9, 12, 18, .........

\(a_2-a_1\) \(=9-3=6,\)

\(a_3-a_2\) \(=12-9=3\)

\(\because\) \(a_3-a_2\) \(\neq a_2-a_1\)

\(\therefore\) 3, 9, 12, 18, ......... will not form an A.P.

+1 vote
by (43.4k points)
edited by

Correct option is A) 2, 4, 6, 8, 

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